Tuesday, September 29, 2015

Yes, I have an old telescope, which rarely gets used, and I was set up trying to align it with the Moon and NOT move it when I tried to put the tablet camera up against the lens. And, of course, I had to wait for the large clouds to move out of the way, by which point, I had to reposition the telescope.

Actually, I had to physically move it. Between trees and neighboring buildings, I had to shift it in the driveway. By the time the Moon was red, I would have had to have put the telescope in one specific spot at the other side of the house. However, by that point, I'd already put it back in the basement, so no Blood Moon pics. Seriously, I had enough just getting what you saw, and I got those by accident. By mistake, I took video. Those are three still images from the video I shot.

Tuesday, September 15, 2015

Seems that I never got around to posting Parts 2, 3, and 4 of the non-Common Core Geometry test. Sorry for the delay.

Here are the questions, with answers and explanations, for the New York State Geometry Regents (not Common Core) exam, Part 2. There were 6 questions, each worth 2 credits. Partial credit may be earned for correct work on a problem without a solution, or for a problem with a solution that contains one computational or conceptual error. All work must be shown. In general, a correct answer without any work is worth 1 credit, unless that answer is given as a choice and an explanation is required.

Part 2

29. The image of after a reflection through the origin is R'S'. If the coordinates of the endpoints of are R(2,-3) and S(5,1), state and label the coordinates of R' and S '.[The use of the set of axes below is optional.]

30. A paper container in the shape of a right circular cone has a radius of 3 inches and a height of 8 inches. Determine and state the number of cubic inches in the volume of the cone, in terms
of π.

The Volume of a Right Circular Cone is found using, V = (1/3)Bh, where B is the area of the base. The area of the Base is πr2, so V = (1/3)πr2h = (1/3)π(3)2(8) = 24π.

31. In isosceles triangle RST shown below, RS = RT, M and N are midpoints of RS and RT, respectively, and MN is drawn. If MN = 3.5 and the perimeter of triangle RST is 25, determine and
state the length of NT.

MN is the midsegment, so ST is twice as long. Since MN = 3.5, ST = 7. RST is isosceles and the perimeter is 25, so x + x + 7 = 25. 2x + 7 = 25, 2x = 18, x = 9.
RS and RT have lengths of 9. N is the midpoint of RT, so NT is 1/2 of RT. NT = 4.5

32. In the diagram below, ABC is equilateral.
Using a compass and straightedge, construct a new equilateral triangle congruent to ABC in
the space below.
[Leave all construction marks.]

Make a point on the lower portion of the page. Call it P. Use the compass to measure the length of AB. Make a little arc. Go back to the point you made and, without changing the compass, make an arc. Make a point on the arc. Call it Q. Use the straightedge to make PQ. Still not changing the compass, make an arc from point P above the middle of the line segment. Make a similar arc from point Q. Where the two arcs intersect, make a point. Call it R. Use the straightedge to make PR and QR. PQR is an equilateral triangle.

About Me

Mr. Burke is a high school math teacher in New York as well as a part-time writer, and a fan of science-fiction/fantasy books and films.
He started making his own math webcomic totally by accident as a way of amusing his students and trying to make them think just a little bit more.
Unless otherwise stated, all math cartoons and other images on this webpage are the creation and property of Mr. Chris Burke and cannot be reused without permission.
Thank you.